Alternative approach: $\sqrt[3]{x}$ is a concave function on $\mathbb{R}^+$, hence for any $n\geq 1$
$$ \sqrt[3]{n^3+n}-\sqrt[3]{n^3-n}=n\left(\sqrt[3]{1+\frac{1}{n^2}}-\sqrt[3]{1-\frac{1}{n^2}}\right)\geq \frac{2}{3n} $$
and the conclusion is straightforward.